Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (T x : p) os << x << " ";
  os << "]" << endl;
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}" << endl;
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const map<Tk, Tv>& p) {
  os << "{ ";
  for (pair<Tk, Tv> x : p) os << x << " ";
  os << "}" << endl;
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const pair<Tk, Tv>& p) {
  os << "{" << p.first << ',' << p.second << "}";
  return os;
}
const int MOD = 1000000000 + 7;
const int INF = 1000000000 + 5;
const int MAX = 200000;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int g[105][105];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &g[i][j]);
    }
  }
  int minMoves = INF;
  vector<int> pr, pc;
  for (int r0 = 0; r0 <= g[0][0]; r0++) {
    vector<int> c(m);
    bool check = true;
    for (int i = 0; i < m; i++) {
      c[i] = g[0][i] - r0;
      if (c[i] < 0) {
        check = false;
        break;
      }
    }
    if (check) {
      vector<int> r(n);
      for (int i = 0; i < n; i++) {
        r[i] = g[i][0] - c[0];
        if (r[i] < 0) {
          check = false;
          break;
        }
      }
      if (check) {
        int moves = 0;
        for (int i = 0; i < n; i++) moves += r[i];
        for (int i = 0; i < m; i++) moves += c[i];
        bool isTrue = true;
        {
          for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
              if (r[i] + c[j] != g[i][j]) {
                isTrue = false;
                break;
              }
            }
            if (!isTrue) break;
          }
        }
        if (isTrue && moves < minMoves) {
          pr = r;
          pc = c;
          minMoves = moves;
        }
      }
    }
  }
  if (minMoves == INF)
    printf("-1\n");
  else {
    printf("%d\n", minMoves);
    for (int i = 0; i < n; i++) {
      while (pr[i]--) printf("row %d\n", i + 1);
    }
    for (int i = 0; i < m; i++) {
      while (pc[i]--) printf("col %d\n", i + 1);
    }
  }
}